\newproblem{lay:2_1_25}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.1.25}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Suppose $A$ is an $m\times n$ matrix and there exist $n\times m$ matrices $C$ and $D$ such that $CA=I_n$ and $AD=I_m$. Prove that $m=n$ and $C=D$. 
	[\textit{Hint}: think of the product $CAD$.]
}{
  % Solution
	Let us compute
	\begin{center}
		$(CA)D=I_nD=D$
	\end{center}
	On the other side, let us compute
	\begin{center}
		$C(AD)=CI_m=C$
	\end{center}
	But we know that matrix multiplication is associative and, consequently, $C=D$.
	
	By Exercise Lay 2.1.23 we know that $A$ cannot have more columns than rows, and by Exercise Lay 2.1.26 we know that $A$ cannot have more rows than columns. Consequently,
	the number of rows and columns must be the same and $m=n$.
}
\useproblem{lay:2_1_25}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
